BPI is equivalent to extending proper Boolean filters

Let $I$ be a proper ideal of $B$. By F2 its quotient map $q:B\to B\mathrm{/}I$ has kernel $I$, so $q(1)\mathrm{\ne }q(0)$ because $1\mathrm{\notin }I$. BPI therefore provides a prime ideal $J$ in $B\mathrm{/}I$. The inverse image $q^{-1}(J)$ contains $I$, is downward and join closed by monotonicity and join preservation, contains $0$ and excludes $1$. If $a\wedge b$ is in it, then $q(a)\wedge q(b)\in J$, so primality of $J$ gives $a$ or $b$ in the inverse image. This proves prime extension.

For a proper filter $F$, let $I=\{\mathrm{\neg }f:f\in F\}$, a proper ideal by F3. Step 1.1 gives a prime ideal $J\supseteq I$. The ultrafilter $U=B\setminus J$ contains $F$: for $f\in F$, $\mathrm{\neg }f\in J$ precludes $f\in J$ because $J$ is proper. Thus BPI also gives filter extension.

Conversely, if prime extension holds, apply it to the proper ideal $\{0\}$ of any nontrivial $B$ to get BPI. If filter extension holds instead, apply it to the proper filter $\{1\}$ and use F3 to take the complement of the resulting ultrafilter. Trivial algebras have neither proper ideals nor proper filters and are excluded by F1. No infinite family of choices is made in any implication. QED.